It is currently 18 Aug 2017, 09:11

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If ab ≠ 0, is |a-b| > |a+b|?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 19 Dec 2015
Posts: 29

Kudos [?]: 68 [2] , given: 11

If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 08:46
2
KUDOS
18
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

61% (02:02) correct 39% (01:23) wrong based on 495 sessions

### HideShow timer Statistics

If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

1) $$ab < 0$$

2) $$a > b$$
[Reveal] Spoiler: OA

Kudos [?]: 68 [2] , given: 11

Manager
Joined: 12 Jan 2016
Posts: 72

Kudos [?]: 23 [0], given: 71

Location: United States
Concentration: Operations, General Management
GMAT 1: 690 Q49 V35
GPA: 3.5
WE: Supply Chain Management (Consumer Electronics)
Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:21
For |a-b| > |a+b|, a and b should have opposite signs. So, if a > 0, b should be less than 0 and vice versa.

1) ab < 0
this proves a and b have opposite signs. So, sufficient.

2) a>b
not sufficient as it does not prove a and b will have opposite signs.
for e.g. a=1, b=-2 --> |3| > |1| --> Yes
a=3, b =2 --> |1| < |5| --> No

Kudos [?]: 23 [0], given: 71

Math Expert
Joined: 02 Sep 2009
Posts: 40927

Kudos [?]: 118623 [8] , given: 12000

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:25
8
KUDOS
Expert's post
11
This post was
BOOKMARKED
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

Square $$|a-b| > |a+b|$$ (we can safely do this since both sides are nonnegative): is $$a^2 - 2ab + b^2 > a^2 +2ab + b^2$$ --> is $$ab < 0$$?

(1) $$ab < 0$$. Directly answers the question. Sufficient.

(2) $$a > b$$. Not sufficient to say whether ab < 0.

_________________

Kudos [?]: 118623 [8] , given: 12000

Intern
Joined: 19 Dec 2015
Posts: 29

Kudos [?]: 68 [0], given: 11

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:31
Bunuel wrote:
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

Square $$|a-b| > |a+b|$$ (we can safely do this since both sides are nonnegative): is $$a^2 - 2ab + b^2 > a^2 +2ab + b^2$$ --> is $$ab < 0$$?

(1) $$ab < 0$$. Directly answers the question. Sufficient.

(2) $$a > b$$. Not sufficient to say whether ab < 0.

Thanks for the reply, Bunuel! Just a quick follow-up: can we assume that both sides are non-negative since ab ≠ 0?

Kudos [?]: 68 [0], given: 11

Math Expert
Joined: 02 Sep 2009
Posts: 40927

Kudos [?]: 118623 [1] , given: 12000

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:35
1
KUDOS
Expert's post
Sallyzodiac wrote:
Bunuel wrote:
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

Square $$|a-b| > |a+b|$$ (we can safely do this since both sides are nonnegative): is $$a^2 - 2ab + b^2 > a^2 +2ab + b^2$$ --> is $$ab < 0$$?

(1) $$ab < 0$$. Directly answers the question. Sufficient.

(2) $$a > b$$. Not sufficient to say whether ab < 0.

Thanks for the reply, Bunuel! Just a quick follow-up: can we assume that both sides are non-negative since ab ≠ 0?

Both sides are nonnegative because both sides are absolute values which cannot be negative.
_________________

Kudos [?]: 118623 [1] , given: 12000

Intern
Joined: 25 Aug 2015
Posts: 5

Kudos [?]: 2 [1] , given: 2

If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:52
1
KUDOS
Hope this helps

Last edited by Z10201Z on 06 May 2016, 09:54, edited 1 time in total.

Kudos [?]: 2 [1] , given: 2

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4780

Kudos [?]: 4930 [2] , given: 109

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

06 May 2016, 09:53
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
Sallyzodiac wrote:
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

1) $$ab < 0$$

2) $$a > b$$

Hi,
Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is $$|a-b| > |a+b|$$?..
We have a number line in which a and b and -b are marked..
the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO
B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) $$ab < 0$$
Both a and b are of opposite sign .. ans YES
Suff

2) $$a > b$$
Nothing ..
both can be same sign or opposite sign ..
the above TWO cases still remain.
.
Insuff
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 4930 [2] , given: 109

VP
Joined: 26 Mar 2013
Posts: 1110

Kudos [?]: 250 [0], given: 148

If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

07 May 2016, 12:36
chetan2u wrote:
Sallyzodiac wrote:
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

1) $$ab < 0$$

2) $$a > b$$

Hi,
Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is $$|a-b| > |a+b|$$?..
We have a number line in which a and b and -b are marked..
the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO
B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) $$ab < 0$$
Both a and b are of opposite sign .. ans YES
Suff

2) $$a > b$$
Nothing ..
both can be same sign or opposite sign ..
the above TWO cases still remain.
.
Insuff

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Kudos [?]: 250 [0], given: 148

Intern
Joined: 25 Aug 2015
Posts: 5

Kudos [?]: 2 [1] , given: 2

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

07 May 2016, 16:04
1
KUDOS
Mo2men wrote:
chetan2u wrote:
Sallyzodiac wrote:
If $$ab ≠ 0$$, is $$|a-b| > |a+b|$$?

1) $$ab < 0$$

2) $$a > b$$

Hi,
Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is $$|a-b| > |a+b|$$?..
We have a number line in which a and b and -b are marked..
the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO
B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) $$ab < 0$$
Both a and b are of opposite sign .. ans YES
Suff

2) $$a > b$$
Nothing ..
both can be same sign or opposite sign ..
the above TWO cases still remain.
.
Insuff

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Take a look at my previous answer. You must analyze four different situations because the two sides of the equation are not independent.

Hope it helps

Best regards

Kudos [?]: 2 [1] , given: 2

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4780

Kudos [?]: 4930 [1] , given: 109

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

07 May 2016, 19:23
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Mo2men wrote:

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Hi,

the problem in your solution is opening of MOD..
yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

$$|a-b|>|a+b|..$$
square both sides..
$$(a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....$$....
so we should find out if ab<0

and statement I tells us that ab<0 .. so suff
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 4930 [1] , given: 109

VP
Joined: 26 Mar 2013
Posts: 1110

Kudos [?]: 250 [0], given: 148

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

07 May 2016, 23:31
chetan2u wrote:
Mo2men wrote:

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Hi,

the problem in your solution is opening of MOD..
yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

$$|a-b|>|a+b|..$$
square both sides..
$$(a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....$$....
so we should find out if ab<0

and statement I tells us that ab<0 .. so suff

Hi chetan2u,

Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality.

1- I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: |X|=|Y|

X=Y or X=-Y or -X=-Y (which is same as X=Y) or -X=Y (Which is the same as X=-Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error.

2- if the question is |a-b|>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know.

I hope one day you can make another post for absolute modulus under inequality.

Kudos [?]: 250 [0], given: 148

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4780

Kudos [?]: 4930 [0], given: 109

Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

08 May 2016, 23:05
Mo2men wrote:
chetan2u wrote:
Mo2men wrote:

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Hi,

the problem in your solution is opening of MOD..
yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

$$|a-b|>|a+b|..$$
square both sides..
$$(a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....$$....
so we should find out if ab<0

and statement I tells us that ab<0 .. so suff

Hi chetan2u,

Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality.

1- I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: |X|=|Y|

X=Y or X=-Y or -X=-Y (which is same as X=Y) or -X=Y (Which is the same as X=-Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error.

2- if the question is |a-b|>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know.

I hope one day you can make another post for absolute modulus under inequality.

hi
lets solve |a-b|>a+b..
this can be straight ..
1) a-b>a+b......2b<0..... b<0..
2) -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..

Or
Critical points..

a is 0, when a is b, so b is our critical point..
this gives us two ranges
a<b or a>=b..
1) a<b...
then a-b will be -ive..
so -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..
2) a>=b..
then a-b will be +ive or 0..
so a-b>a+b......2b<0..... b<0..

Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 4930 [0], given: 109

VP
Joined: 26 Mar 2013
Posts: 1110

Kudos [?]: 250 [0], given: 148

If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

08 May 2016, 23:39
chetan2u wrote:
hi
lets solve |a-b|>a+b..
this can be straight ..
1) a-b>a+b......2b<0..... b<0..
2) -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..

Or
Critical points..

a is 0, when a is b, so b is our critical point..
this gives us two ranges
a<b or a>=b..
1) a<b...
then a-b will be -ive..
so -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..
2) a>=b..
then a-b will be +ive or 0..
so a-b>a+b......2b<0..... b<0..

Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method

I think there is typo. it should: a-b=0, when a is b

Kudos [?]: 250 [0], given: 148

Manager
Joined: 17 Nov 2014
Posts: 52

Kudos [?]: 3 [0], given: 141

Location: India
Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

15 May 2016, 04:27
Rewording the question , |a-b| > |a+b|, a and b should have opposite signs.

Statement 1 exactly points out that :- Hence sufficient.

Statement 2 Insufficient

Kudos [?]: 3 [0], given: 141

Math Forum Moderator
Joined: 13 Apr 2015
Posts: 1504

Kudos [?]: 979 [0], given: 871

Location: India
Concentration: Strategy, General Management
WE: Information Technology (Consulting)
Re: If ab ≠ 0, is |a-b| > |a+b| ? [#permalink]

### Show Tags

25 Jun 2016, 10:42
Is |a - b| > |a + b| ?
Square on both sides --> Is a^2 + b^2 - 2ab > a^2 + b^2 + 2ab --> Is 4ab < 0 ?
Is ab < 0?

St1: ab < 0 --> Directly answers the question. Sufficient

St2: a > b --> Clearly Insufficient.
If a = 2 and b = 1 then ab > 0
If a = 2 and b = -1 then ab < 0

Kudos [?]: 979 [0], given: 871

Manager
Joined: 20 Mar 2015
Posts: 68

Kudos [?]: 23 [0], given: 9

If ab ≠ 0, is |a-b| > |a+b| ? [#permalink]

### Show Tags

25 Jun 2016, 11:01
shiv242003 wrote:
If ab ≠ 0, is |a-b| > |a+b| ?

(1) ab < 0
(2) a > b

Thanks ,
Shiv.

According to statement 1. a*b is negative. So, either a or b is negative. Both of them can't be negative. by question a*b doesn't equals to zero.
take testing numbers.
a = 1 or -1
b = -1 or 1
if a = 1 and b =-1 then;
|1 -(-1)| > |1 -1|
or, |1+1| > |0|
or, 2>0. true.

also, check for a= -1 and b =1
|-1-(1)| > |-1+1|
or, |-1-1| > |0|
or, 2>0. true.
so Statement 1 is sufficient.

According to statement 2, a>b.
take testing points.
a=-1 or 2
b= -2 or -1
if a= -1 and b = -2 then,
|-1 - (-2)| > |-1+(-2)|
or, |-1+2|> |-1-2|
or, |1|>|-3|
or 1>3 false.

If a= 2 and b = -1
|2-(-1)|>|2-1|
or,|3|>|1|
or, 3>1 true.
So, statement 2 is not sufficient to answer the question. Statement 2 gives either true or false. Whereas, Statement 1 gives true only.

I was waiting for an easy solution and Vyshak gives that.

Kudos [?]: 23 [0], given: 9

Manager
Joined: 25 Jun 2016
Posts: 61

Kudos [?]: 58 [1] , given: 4

GMAT 1: 780 Q51 V46
Re: If ab ≠ 0, is |a-b| > |a+b| ? [#permalink]

### Show Tags

25 Jun 2016, 11:12
1
KUDOS
You can think of absolute value as distance from zero. So we need enough info to definitely determine whether a - b is further from zero than the sum a + b.

Statement one tells us that a and b have different signs (that's the only way that the product will be negative).

For the difference a - b, this means that the numbers end up 'working together' to get away from zero; if you take a positive number away from a negative number, you go further down, further away from zero (e.g. -10 - 2 = -12). If you take a negative number away from a positive number you go further up (e.g. 10 - - 2 = 12) Takeaway: The difference of numbers with opposite signs is always further from zero than either of the original numbers.

For the sum a + b, statement one means that the numbers end up 'working in opposite directions.' That is, when we add numbers of opposite signs, one cancels out part of the other's distance to zero. Think of 10 + - 2 and -2 + 10. In both cases we end up closer to zero than the furthest term was to begin with. Takeaway: the sum of two numbers with opposite signs will be closer to zero than the original number of greatest magnitude (the term that was furthest from zero to begin with).

So if we know two numbers have opposite signs, we know the difference of those numbers ends up further from zero than the sum. Sufficient.

statement 2 doesn't tell us anything about the signs, so we might be suspicious right from the start. And sure enough, if all we know is that a is to the right of b, then we might have a couple of positives, (such as a = 3, and b = 2) in which case the sum is further from zero than the difference. But we could also have a positive and a negative (such as a = 3, and b = - 2), in which case the difference is further from zero than the sum. Insufficient.

Kudos [?]: 58 [1] , given: 4

Intern
Joined: 06 Jul 2016
Posts: 44

Kudos [?]: 38 [0], given: 2

Location: India
Concentration: Operations, General Management
GMAT 1: 610 Q49 V25
WE: General Management (Energy and Utilities)
Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

26 Oct 2016, 11:33
karant wrote:
If ab is not equal to 0, is |a-b| > |a+b| ?

(1) ab < 0

(2) a > b

Need help on this official question.

We can square both sides of '|a-b| > |a+b|' and we will get 4ab<0.

1. Gives us the ab<0. Bingo. That is what required to get the above expression true. Sufficient.

2. a>b. Clearly not sufficent as A and be could be 3 and 2 respectively or -2 and -3. Insufficient.

Kudos [?]: 38 [0], given: 2

Manager
Joined: 26 Apr 2012
Posts: 94

Kudos [?]: 16 [0], given: 75

Location: India
Concentration: Entrepreneurship, General Management
GMAT 1: 640 Q48 V29
GMAT 2: 660 Q45 V35
GMAT 3: 680 Q48 V35
GPA: 2.8
WE: Information Technology (Computer Software)
Re: If ab ≠ 0, is |a-b| > |a+b|? [#permalink]

### Show Tags

26 Oct 2016, 12:13
1
This post was
BOOKMARKED
Hello Bunuel, chetan2u

Thanks for the solution. I got the point.

However, I would like to know how this problem will be solved algebrically. Can you please explain the solution by opening the modulus on both the side ?

There will be following two condition if we open modulus:

a-b > a+b
gives, b <0

and

a-b > -(a+b)
gives , a>0

What would be the next steps now ? How to interpret. Please explain.

Thanks

Kudos [?]: 16 [0], given: 75

Senior Manager
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 366 [0], given: 40

GPA: 3.98
Re: If ab ≠ 0, is |a-b|>|a+b|? [#permalink]

### Show Tags

07 Jan 2017, 08:35
nayanparikh wrote:
If ab ≠ 0, is |a-b|>|a+b|?

1) ab < 0
2) a > b

Can any one please provide explanation on how to solve these type of questions ?
@Bunnel ?

1) ab < 0

Means that there are two possibilities: either a>0 and b<0 or a<0 and b>0. Let's look at htose two seperately.

a>0, b<0. Let's denote b=-y (y - some positive integer). -y<0, y>0.

|(a - (-y))| > |a - y|

|a + y| > |a - y|

On the left side we have sum of two positive integers so we can pull off our modulus. On the right side we have difference o two positive integers which will always be lesser than their sum.

a<0, b>0. Same approach. a=-x (x some positive integer). -x<0, x>0

|-x - b| > |-x + b|

| - (x +b)| > |b - x|

x + b > |b - x|

Again we have sum of two positive integers on the left and their diffrence on th right. Sufficient.

2) a > b

We have three possible scenarious.

a) a>0, b>0
b) a<0, b<0
c) a>0, b<0

a) To eveade confusion it would be impler to use particular numbers here.

a = 30, b = 10

|30 - 10| ... |30 + 10| ----> 20 < 40

b) a = -10 b = -30

|-10 - (-30)| ... | -10 - 30|

| -10 + 30| ... | - (10 + 30)|

20 < 40

c) In c we have same case as in (1)

a=10, b=-30

|10 - (-30)| ... |10 - 30|

40 > 20 Insufficient

I hope my approach wasn't too vague. May be our experts will give you more details.

Kudos [?]: 366 [0], given: 40

Re: If ab ≠ 0, is |a-b|>|a+b|?   [#permalink] 07 Jan 2017, 08:35

Go to page    1   2    Next  [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
11 Is 0 > ab? 11 21 Aug 2016, 06:16
96 If ab ≠ 0, is ab > a/b ? 27 02 Feb 2017, 17:49
2 Is a/b > 0 ? 6 13 Mar 2016, 08:41
6 Is ab > 0? 6 22 Apr 2017, 09:28
2 Is a/b > 0? 3 18 Jul 2016, 04:13
Display posts from previous: Sort by